A360010 First part of the n-th weakly decreasing triple of positive integers sorted lexicographically. Each n > 0 is repeated A000217(n) times.

1, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8

Offset

1,2

Links

Table of n, a(n) for n=1..87.

Formula

a(n) = A056556(n) + 1 = A331195(3n) + 1.

Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/8 + log(2)/4. - Amiram Eldar, Feb 18 2024

a(n) = m+1 if n>binomial(m+2,3) and a(n) = m otherwise where m = floor((6n)^(1/3)). - Chai Wah Wu, Nov 04 2024

Example

Triples begin: (1,1,1), (2,1,1), (2,2,1), (2,2,2), (3,1,1), (3,2,1), (3,2,2), (3,3,1), (3,3,2), (3,3,3), ...

Mathematica

nn=9; First/@Select[Tuples[Range[nn], 3], GreaterEqual@@#&]

Prog

(Python)
from math import comb
from sympy import integer_nthroot
def A360010(n): return (m:=integer_nthroot(6*n, 3)[0])+(n>comb(m+2, 3)) # Chai Wah Wu, Nov 04 2024

Crossrefs

For pairs instead of triples we have A002024.

Positions of first appearances are A050407(n+2) = A000292(n)+1.

The zero-based version is A056556.

The triples have sums A070770.

The second instead of first part is A194848.

The third instead of first part is A333516.

Concatenating all the triples gives A360240.

Cf. A000217, A003056, A056557, A056558, A069905, A158842, A331195.

Sequence in context: A346425 A029836 A004257 * A276611 A238598 A156684

Adjacent sequences: A360007 A360008 A360009 * A360011 A360012 A360013

Keyword

nonn

Author

Gus Wiseman, Feb 11 2023

Status

approved